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全同胞对中肠易激综合征评分概率分布的方程推导

Equation Derivation of the Probability Distribution of IBS Score among Full Sibling Pairs.

作者信息

Zhao Q, Li R, Jin Y Z, Sun H Y, Zhao S M

机构信息

Shanghai Cubicise Biotechnology Co. Ltd, Shanghai 201114, China.

Department of Forensic Medicine, Zhongshan School of Medicine, Sun Yat-sen University, Guangzhou 510080, China.

出版信息

Fa Yi Xue Za Zhi. 2019 Dec;35(6):657-661. doi: 10.12116/j.issn.1004-5619.2019.06.003. Epub 2019 Dec 25.

DOI:10.12116/j.issn.1004-5619.2019.06.003
PMID:31970950
Abstract

Objective To derive the general equation of the probability distribution of identity by state (IBS) score among biological full sibling pairs by calculating STR allele frequency. Methods Based on the Mendelian genetics law and the hypothesis that parents of biological full siblings (FS) were unrelated individuals, the score and corresponding probability of different genotype combinations in the offspring when unrelated individuals of different genotype combinations give birth to two offsprings were derived. Results Given (=1, 2, …, ) as the frequency of the ith allele of a STR locus, the probability of sharing 2 alleles (), 1 allele () or 0 allele () with biological full sibling pairs on the locus can be respectively expressed as follows: (see the text). The sum of , and must be 1. As for the multiple genotyping system that contained STR loci, scores between biological full sibling pairs conform to binomial distribution: ~B(2, 1). The population rate 1, can be given by the formula: (see the text). Conclusion The alternative hypothesis in biological full sibling testing is that two appraised individuals are biological full siblings. The probability of the corresponding alternative hypothesis of any STR locus combination or score can be directly calculated by the equations presented in this study, and the calculation results are the basis for explanations of the evidence.

摘要

目的 通过计算STR等位基因频率推导生物学全同胞对间状态一致性(IBS)评分概率分布的通用方程。方法 基于孟德尔遗传定律及生物学全同胞(FS)的父母为无关个体这一假设,推导不同基因型组合的无关个体生育两个后代时后代中不同基因型组合的IBS评分及相应概率。结果 设STR基因座第i个等位基因的频率为 (=1, 2, …, ),在该基因座上与生物学全同胞对共享2个等位基因()、1个等位基因()或0个等位基因()的概率可分别表示如下:(见原文)。 、 和 的和必为1。对于包含 个STR基因座的多重基因分型系统,生物学全同胞对间的IBS评分符合二项分布:~B(2, 1)。总体率1可由公式给出:(见原文)。结论 生物学全同胞检测中的备择假设是两个被评估个体为生物学全同胞。本研究给出的方程可直接计算任何STR基因座组合或IBS评分相应备择假设的概率,计算结果是证据解释的依据。

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